How do you solve ${log}_{2} (x + 2) - {log}_{2} (x - 5) = 3$ $log_2 (x+2) - log_2 (x-5) = 3$ ?

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Answer 1 · 2016-01-02T15:48:10

Unify the logarithms and cancel them out with ${log}_{2} 2^{3}$ $log_(2)2^3$

$x = 6$ $x=6$

${log}_{2} (x + 2) + {log}_{2} (x - 5) = 3$ $log_(2)(x+2)+log_(2)(x-5)=3$

Property $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$

${log}_{2} (\frac{x + 2}{x - 5}) = 3$ $log_(2)((x+2)/(x-5))=3$

Property $a = {log}_{b} a^{b}$ $a=log_(b)a^b$

${log}_{2} (\frac{x + 2}{x - 5}) = {log}_{2} 2^{3}$ $log_(2)((x+2)/(x-5))=log_(2)2^3$

Since ${log}_{x}$ $log_x$ is a 1-1 function for $x > 0$ $x>0$ and $x \neq 1$ $x!=1$ , the logarithms can be ruled out:

$\frac{x + 2}{x - 5} = 2^{3}$ $(x+2)/(x-5)=2^3$

$\frac{x + 2}{x - 5} = 8$ $(x+2)/(x-5)=8$

$x + 2 = 8 (x - 5)$ $x+2=8(x-5)$

$x + 2 = 8 x - 8 \cdot 5$ $x+2=8x-8*5$

$7 x = 42$ $7x=42$

$x = \frac{42}{7}$ $x=42/7$

$x = 6$ $x=6$

How do you solve log2(x+2)−log2(x−5)=3log_2 (x+2) - log_2 (x-5) = 3?